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charle [14.2K]
3 years ago
13

Consider the reaction: Mg(OH)₂ (s) ⇌ Mg²⁺ (aq) + 2 OH⁻ (aq)

Chemistry
1 answer:
balandron [24]3 years ago
8 0

The value of Kc for the equilibrium is 0.150 mole² / litre ²

<u>Explanation:</u>

<u>Given:</u>

An equilibrium mixture in an 1.00 L vessel contains 5.30 moles of  

Mg(OH )₂  0.800 moles of Mg²⁺ and 0.0010 moles OH₋  

We have to find the value of Kc

  • Step 1: Find  the equilibrium Concentration.
  • Step 2: Substitute the values in the equation.
  • Step 3: Find the value of Kc.
  • I have attached the document for the detailed explanation

The value of Kc for the equilibrium is 0.150 mole² / litre ²

Download pdf
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Toluene (C6H5CH3 ), an organic compound often used as a solvent in paints, is mixed with a similar organic compound, benzene (C6
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Explanation:

The given data is as follows.

 Weight of solute = 75.8 g,   Molecular weight of solute (toulene) = 92.13 g/mol,    volume = 200 ml

  • Therefore, molarity of toulene is calculated as follows.

      Molarity = \frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}

                    = \frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}

                    = 4.11 M

Hence, molarity of toulene is 4.11 M.

  • As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

   Molality = \frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}

             = \frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}

             = 8.6 m

Hence, molality of given toulene solution is 8.6 m.

  • Now, calculate the number of moles of toulene as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{75.8 g}{92.13 g/mol}

                             = 0.8227 mol

Now, no. of moles of benzene will be as follows.

     No. of moles = \frac{mass}{\text{molar mass}}

                             = \frac{95.6 g}{78.11 g/mol}

                             = 1.2239 mol

Hence, the mole fraction of toulene is as follows.

         Mole fraction = \frac{\text{moles of toulene}}{\text{total moles}}

                             = \frac{0.8227 mol}{(0.8227 + 1.2239) mol}

                             = 0.402

Hence, mole fraction of toulene is 0.402.

  • As density of given solution is 0.857 g/cm^{3} so, we will calculate the mass of solution as follows.

         Density = \frac{mass}{volume}

     0.857 g/cm^{3} = \frac{mass}{200 ml}      (As 1 cm^{3} = 1 g)

                      mass = 171.4 g

Therefore, calculate the mass percent of toulene as follows.

      Mass % = \frac{\text{mass of solute}}{\text{mass of solution}} \times 100

                   = \frac{75.8 g}{171.4 g} \times 100

                   = 44.22%

Therefore, mass percent of toulene is 44.22%.

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